Physics 1202: Lecture 5Today’s Agenda
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– HW assignments, solutions etc.
• Homework #2:Homework #2:– On Masterphysics today: due Friday next weekOn Masterphysics today: due Friday next week
– Go to masteringphysics.com
• Labs: Begin next week
Today’s Topic :• End of Chapter 16: Electric potential
– Equipotentials and Conductors
– Electric Potential Energy
» of Charge in External Electric Field
• Capacitors:Electrostatic energy– Definition and concept
– Capacitors in parallel and in series
– Energy stored
– Dielectrics
Electric Potential
qA
C
B
rA
Br
path independence equipotentials
R
R
R r
VQ
4 rQ
4 R
Electric Potential• By analogy with the electric field
• Defined using a test charge q0
• We define a potential V due to a charge q
– Using potential energy of a charge q and a test charge q0
Electric Potential Energy• The Coulomb force is a CONSERVATIVE force (i.e. the work
done by it on a particle which moves around a closed path returning to its initial position is ZERO.)
• Therefore, a particle moving under the influence of the Coulomb force is said to have an electric potential energy defined by:
• The total energy (kinetic + electric potential) is then conserved for a charged particle moving under the influence of the Coulomb force.
this “q” is the ‘test charge” in other examples...
E from V?• We can obtain the electric field E from the potential V by
inverting our previous relation between E and V:
• We found
++
++
++
++
++
++
++
++
++
++
++
++
++
- -
- -
- -
- -
- -
- -
- -
- -
- -
- -
- -
- -
- -
+
F
• In general true for all direction
Equipotentials
• GENERAL PROPERTY: – The Electric Field is always perpendicular to an
Equipotential Surface.
• Why??
Dipole Equipotentials
Defined as: The locus of points with the same potential.
• Example: for a point charge, the equipotentials are spheres centered on the charge.
Along the surface, there is NO change in V (it’s an equipotential!)
So, there is NO E component along the surface either… E must therefore be normal to surface
• ClaimThe surface of a conductor is always an equipotential surface(in fact, the entire conductor is an equipotential)
• Why??
If surface were not equipotential, there would be an Electric Field component parallel to the surface and the charges would move!!
• NotePositive charges move from regions of higher potential to lower potential (move from high potential energy to lower PE).
Equilibrium means charges rearrange so potentials equal.
Conductors
+ +
+ +
+ +
+ + +
+ + +
+ +
A Point Charge Near Conducting Plane
+
a
q
- - -- -- - - -- - - - -- --- ---------------- --- -- ---- --------V=0
A Point Charge Near Conducting Plane
+
-
a
q
The magnitude of the force is
The test charge is attracted to a conducting plane
Image Charge
Equipotential Example• Field lines more closely
spaced near end with most curvature .
• Field lines to surface near the surface (since surface is equipotential).
• Equipotentials have similar shape as surface near the surface.
• Equipotentials will look more circular (spherical) at large r.
Definitions & Examples
d
A
- - - - -
+ + + +a
b L
a
b
a b
Capacitance• A capacitor is a device whose purpose is to store electrical
energy which can then be released in a controlled manner during a short period of time.
• A capacitor consists of 2 spatially separated conductors which can be charged to +Q and -Q respectively.
• The capacitance is defined as the ratio of the charge on one conductor of the capacitor to the potential difference between the conductors.
• Is this a "good" definition? Does the capacitance belong only to the capacitor, independent of the charge and voltage ?
+ -
• Calculate the capacitance. We assume +, - charge densities on each plate with potential difference V:
Example:Parallel Plate Capacitor
d
A
- - - - -
+ + + +
• Need Q:
• Need V: recall
where x = dor
so
Recall:Two infinite planes• Same charge but opposite
• Fields of both planes cancel out outside
• They add up inside
++++++++++++++++++++++++++
- - - - - - - - - - - - - - - - - - - - - - - - - -
Perfect to store energy !
Example:Parallel Plate Capacitor
d
A
- - - - -
+ + + +
• Calculate the capacitance:
Assume +Q,-Q on plates with potential difference V.
• As hoped for, the capacitance of this capacitor depends only on its geometry (A,d).
Dimensions of capacitance
• C = Q/V => [C] = F(arad) = C/V = [Q/V]
• Example: Two plates, A = 10cm x 10cm d = 1cm apart
=> C = A0/d = = 0.01m2/0.01m * 8.852e-12 C2/Jm = 8.852e-12 F
Lecture 5 – ACT 1
d
A
- - - - -
+ + + +• Suppose the capacitor shown here is charged to Q and then the battery disconnected.
• Now suppose I pull the plates further apart so that the final separation is d1. d1 > d d1
A
- - - - -
+ + + +
If the initial capacitance is C0 and the final capacitance is C1, is …
A) C1 > C0 B) C1 = C0 C) C1 < C0
• Can we define the capacitance of a single isolated sphere ?
• The sphere has the ability to store a certain amount of charge at a given voltage (versus V=0 at infinity)
Example :Isolated Sphere
• Need V: V = 0
VR = keQ/R
• So, C = R/ke
+
+
+
+
Capacitors in Parallel
• Find “equivalent” capacitance C in the sense that no measurement at a,b could distinguish the above two situations.
V
a
b
Q2Q1 V
a
b
Q
Parallel Combination:
Equivalent Capacitor:
Total charge: Q = Q1 + Q2
C = C1 + C2
Capacitors in Series
• Find “equivalent” capacitance C in the sense that no measurement at a,b could distinguish the above two situations.
• The charge on C1 must be the same as the charge on C2 since applying a potential difference across ab cannot produce a net charge on the inner plates of C1 and C2 .
a b+Q -Q
a b+Q -Q
RHS:
LHS:
Examples:Combinations of Capacitors
a
b
a b
• How do we start??
• Recognize C3 is in series with the parallel combination on C1 and C2. i.e.
Energy of a Capacitor• How much energy is stored in a charged capacitor?
– Calculate the work provided (usually by a battery) to charge a capacitor to +/- Q:
Calculate incremental work W needed to add charge q to capacitor at voltage V: - +
• But W is also the change in potential energy U
q
Vq
Vq=q/CV
• The total U to charge to Q is shaded triangle:
• In terms of the voltage V:
Lecture 5 – ACT 2
d
A
- - - - -
+ + + +
d1
A
- - - - -
+ + + +
The same capacitor as last time.
The capacitor is charged to Q and then the battery disconnected.
Then I pull the plates further apart so that the final separation is d1. d1 > d
If the initial energy is U0 and the final
capacitance is U1, is …
A) U1 > U0 B) U1 = U0 C) U1 < U0
Summary
d
A
- - - - -
+ + + +• Suppose the capacitor shown here is charged to Q and then the battery disconnected.
• Now suppose I pull the plates further apart so that the final separation is d1.
• How do the quantities Q, W, C, V, E change?
• How much do these quantities change?.. exercise for student!!
• Q:• W:• C:• V:• E:
remains the same.. no way for charge to leave.
increases.. add energy to system by separating
decreases.. since energy , but Q remains sameincreases.. since C , but Q remains sameremains the same.. depends only on chg density
answers:
Where is the Energy Stored?• Claim: energy is stored in the Electric field itself. Think of the
energy needed to charge the capacitor as being the energy needed to create the field.
• The Electric field is given by:
• The energy density u in the field is given by:
• To calculate the energy density in the field, first consider the constant field generated by a parallel plate capacitor:
Units: J/m3
Dielectrics• Empirical observation:
Inserting a non-conducting material between the plates of a capacitor changes the VALUE of the capacitance.
• Definition:
The dielectric constant of a material is the ratio of the capacitance when filled with the dielectric to that without it. i.e.
– values are always > 1 (e.g., glass = 5.6; water = 78)
– They INCREASE the capacitance of a capacitor (generally good, since it is hard to make “big” capacitors
– They permit more energy to be stored on a given capacitor than otherwise with vacuum (i.e., air)
Parallel Plate Example +++++++++++++
- - - - - - - - - - - - -
• Charge a parallel plate capacitor filled with vacuum (air) to potential difference V0.
• An amount of charge Q = C V0 is deposited on each plate.
+++++++++++++
- - - - - - - - - - - - -
• Now insert material with dielectric constant .
– Charge Q remains constant
+
- +
-
+
- +
-
+
-
+
-+
-
– So…, C = C0
Voltage decreases from V0 to
Electric field decreases also: